How to Show u(x,y) and v(x,y) are Constant Throughout D?

tylerc1991
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Homework Statement



Suppose v is a harmonic conjugate of u in a domain D, and that u is a harmonic conjugate of v in D. Show how it follows that u(x,y) and v(x,y) are constant throughout D.

The Attempt at a Solution



since u is a harmonic conjugate of v, u_xx + u_yy = 0
also, since v is a harmonic conjugate of u, v_xx + v_yy = 0

u_x = v_y => u_xx = v_yx
u_y = -v_x => u_yy = -v_xy

I think that I am going in circles here. Can someone lend a helping hand with this problem? Thank you so much!
 
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Hint: in one direction u_x=v_y and u_y=(-v_x). In the other direction v_x=u_y and v_y=(-u_x). Put them together.
 
what do you mean in one direction?

do you mean that since u is a harmonic conjugate then that is one direction, and since v is a harmonic conjugate then that is the other direction?
 
tylerc1991 said:
what do you mean in one direction?

do you mean that since u is a harmonic conjugate then that is one direction, and since v is a harmonic conjugate then that is the other direction?

Yes, that's exactly what I mean.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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